d. which of these transitions (6 → 2 or 6 → 1) produces light that is (just barely) within the visible…

d. which of these transitions (6 → 2 or 6 → 1) produces light that is (just barely) within the visible spectrum? in what part of the spectrum is the light from the other transition? e. consider an electron making the transition 7 → 2. without doing specific calculations, predict how the wavelength of the light from the 7 → 2 transition would compare to the wavelength for the 6 → 2 transition. in what part of the spectrum would the light be for the 7 → 2 transition?

d. which of these transitions (6 → 2 or 6 → 1) produces light that is (just barely) within the visible spectrum? in what part of the spectrum is the light from the other transition? e. consider an electron making the transition 7 → 2. without doing specific calculations, predict how the wavelength of the light from the 7 → 2 transition would compare to the wavelength for the 6 → 2 transition. in what part of the spectrum would the light be for the 7 → 2 transition?

Answer

Explanation:

Step1: Recall spectral series rules

In hydrogen - like atoms, transitions ending at $n = 2$ are part of the Balmer series (visible light), and transitions ending at $n=1$ are part of the Lyman series (ultraviolet).

Step2: Analyze transition $6\rightarrow2$

The $6\rightarrow2$ transition is part of the Balmer series. The energy difference $\Delta E=E_6 - E_2$. According to $E =-\frac{13.6}{n^{2}}\text{ eV}$, the emitted photon has energy in the visible - light range.

Step3: Analyze transition $6\rightarrow1$

The $6\rightarrow1$ transition is part of the Lyman series. The energy difference $\Delta E'=E_6 - E_1$ is larger than $\Delta E$. Since $E = h\nu=\frac{hc}{\lambda}$, a larger energy difference means a shorter wavelength. The Lyman - series photons are in the ultraviolet part of the spectrum.

Step4: Compare $7\rightarrow2$ and $6\rightarrow2$ transitions

The energy difference for the $7\rightarrow2$ transition is larger than that for the $6\rightarrow2$ transition because $E_7<E_6$ and the final state is the same ($n = 2$). Using $E=\frac{hc}{\lambda}$, a larger energy difference corresponds to a shorter wavelength. Since the $7\rightarrow2$ transition is also part of the Balmer series, it is in the visible part of the spectrum.

Answer:

  • The $6\rightarrow2$ transition produces visible light (Balmer series). The $6\rightarrow1$ transition produces ultraviolet light (Lyman series).
  • The wavelength of the light from the $7\rightarrow2$ transition is shorter than that from the $6\rightarrow2$ transition. The $7\rightarrow2$ transition is in the visible part of the spectrum (Balmer series).